KingGeorge
  • KingGeorge
[SOLVED] Here's a cute predicate logic problem. Prove or disprove: \[\forall x \,\exists y \,\forall z\, \exists w \;\;\text{such that} \;\;x+y+z+w=xyzw\]Assume that \(x, y, z, w \in \mathbb{C}\).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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experimentX
  • experimentX
is xyzw a combination of digit or multiplication of digit??
KingGeorge
  • KingGeorge
multiplication
experimentX
  • experimentX
lol .. didn't realize they were complex

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More answers

KingGeorge
  • KingGeorge
It doesn't really matter. I just put that specification on there so that know one tried to change the problem so that addition and multiplication had different meanings.
KingGeorge
  • KingGeorge
Just prove/disprove it for \(\mathbb{R}\) and it will become clear why it doesn't matter if they're complex numbers.
KingGeorge
  • KingGeorge
Hint below in white LaTeX\[\color{white}{\text{There is a certain value for y that you should choose to prove it.}}\]
experimentX
  • experimentX
\[ \color{white}{\text{LOL i didn't know that!!.}} \]
KingGeorge
  • KingGeorge
Another, probably more helpful hint, below in white LaTeX\[\color{white}{\text{What value of y can you choose so that xyzw is constant?}}\]
anonymous
  • anonymous
Choose y=0, then choose w=-(x+z).
KingGeorge
  • KingGeorge
Bingo.
experimentX
  • experimentX
But ∃y doesn't mean that y is rather a variable such that we have to fit constraint ??
KingGeorge
  • KingGeorge
\(\exists\, y\) means that we can choose a specific value for y. It's a variable, but one that we have complete control over.
experimentX
  • experimentX
then it must be false ... i guess so!!
KingGeorge
  • KingGeorge
The "there exists" only means that at least one possible value of y where this is true. One value that we know works is \(y=0\). This means that there does exists at least one value for \(y\) that makes the statement true, so we have proven that the original statement is true.
experimentX
  • experimentX
does't y=0 make statement untrue?? xyzw=0, for y=0
anonymous
  • anonymous
Right, both the sum and the multiplication end up equaling zero because you set w=-(x+z).
KingGeorge
  • KingGeorge
If \(y\) is 0, and \(x=-(x+z)\) we have \[x+0+z+(-(x+z))=x+z-x-z=0\]and \[x\cdot0\cdot z\cdot w=0\]So that means that \[x+y+z+w=xyzw\]
experimentX
  • experimentX
Oh ... stupid me ... i never considered the negative values!!
KingGeorge
  • KingGeorge
Hope everyone enjoyed this little (unintuitive) problem. Good night.
experimentX
  • experimentX
thanks!! i'll try to think broader way next time!!
KingGeorge
  • KingGeorge
I'll try to get a problem just for you next time then. Any preference about the subject?

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